# Firing a bullet into a bock suspended by a string Help!

Hey guys apparently the answer to this is 491 m/s, but i keep getting 34m/s by using a method whereby i find out the height reached by the block then using kinetic energy to potential energy... doesn't seem to work tho! Please help!

A 50.0 g bullet is fired into a stationary 10.0 kg block suspended on a light inextensible wire of length 1.3 m. If the bullet becomes fully embedded in the block, and the bullet-block system reaches a maximum angle of 40.0◦, find the initial velocity of the bullet.

##\delta h## is given by 1.3-1.3cos(40), and ##g\delta h = \frac{1}{2} v_i^2## (conservation of energy, where ##v_i## is the initial velocity of the block+bullet). The total momentum before and after the collision stays the same as well, so you should be able to figure out why the velocity of the bullet had a magnitude of approximately 491 m/s before the collision.

Bystander
Homework Helper
Gold Member
Show your work, and it will make it easier to see what's going on.

##\delta h## is given by 1.3-1.3cos(40), and ##g\delta h = \frac{1}{2} v_i^2## (conservation of energy, where ##v_i## is the initial velocity of the block+bullet). The total momentum before and after the collision stays the same as well, so you should be able to figure out why the velocity of the bullet had a magnitude of approximately 491 m/s before the collision.
I am pretty sure i did this method, and ended up with a result of 34m/s, not sure that method can work for some reason

@StonedPhysicist The method worked just fine for me. Why don't you show your working? As Bystander said, it would make things much easier.

using v=√((2(m+M)gh)/m) , where m=0.05 kg M=10kg g=9.81 and h = 0.3...m i get 34m/s???

Why are you taking a square root for m as well?

Why are you taking a square root for m as well?
i am just rearranging (m+M)gh=1/2 mv2

The kinetic energy should be given by ##\frac{1}{2} (M+m) v^2##. The gain in PE of the block+bullet equals to the KE the block+bullet possessed right after collision. The equation resolves to what I previously provided. So applying conservation of momentum, we get ##mv_1=Mv_2## , so ##v_1=\frac{M}{m} \sqrt{2g\delta h}## .

The kinetic energy should be given by ##\frac{1}{2} (M+m) v^2##. The gain in PE of the block+bullet equals to the KE the block+bullet possessed right after collision. The equation resolves to what I previously provided. So applying conservation of momentum, we get ##mv_1=Mv_2## , so ##v_1=\frac{M}{m} \sqrt{2g\delta h}## .
ah I see now! thankyou!!

haruspex